2+2, you guys were helpfulllll.<br> : )

1. Prove or give a counterexample for the following statements: a) If ff: AA → BB is an injective function and bb ∈ BB, then |ff

Fantom [35]

Answer:

a) False. A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1

b) True

c) True

d) B = {1}, A = N, f: N ⇒ {1}, f(x) = 1

Step-by-step explanation:

a) lets use A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1. Here f is injective but 2 is an element of b and |f−¹({b})| = 0., not 1. This statement is False.

b) This is True. If A were finite, then it can only be bijective with another finite set with equal cardinal, therefore, B should be finite (and with equal cardinal). If A were not finite but countable, then there should exist a bijection g: N ⇒ A, where N is the set of natural numbers. Note that f o g : N ⇒ B is a bijection because it is composition of bijections. This, B should be countable. This statement is True.

c) This is true, if f were surjective, then for every element of B there should exist an element a in A such that f(a) = b. This means that f−¹({b}) has positive cardinal for each element b from B. since f⁻¹(b) ∩ f⁻¹(b') = ∅ for different elements b and b' (because an element of A cant return two different values with f). Therefore, each element of B can be assigned to a subset of A (f⁻¹(b)), with cardinal at least 1, this means that |B| ≤ |A|, and as a consequence, B is finite.

b) This is false, B = {1} is finite, A = N is infinite, however if f: N ⇒ {1}, f(x) = 1 for any natural number x, then f is surjective despite A not being finite.

4 0

3 years ago

Pls look at pic! if u help i’ll give brainliest!!

TiliK225 [7]

Step-by-step explanation:

This is a right triangle with two congruent angles. so this is a isoceles right triangle or called a

45-45-90 triangle. This means that the legs are equal to each other so

$3(x + 2) = 24$

$3x + 6 = 24$

$3x = 18$

$x = 6$

4 0

3 years ago

Convert the fraction to a decimal.

kirza4 [7]

53/100
163/20
5 3/100
8 3/20

6 0

3 years ago

Read 2 more answers

A population of values has a normal distribution with μ = 149.8 μ=149.8 and σ = 68.2 σ=68.2. You intend to draw a random sample

Len [333]

Answer:

There is a 49.20% probability that a single randomly selected value is less than 148.3.

There is a 38.21% probability that a sample of size n = 186 n=186 is randomly selected with a mean less than 148.3.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

A population of values has a normal distribution with $\mu = 149.8$ and $σ=68.2$ .